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a) If $$f(x) = \frac{2x-8}{x^2 -2x - 3} \qquad\text{ and }\qquad g(x) = \frac{3x+9}{2x-4}$$ find the sum of the values of x where the vertical asymptotes of f(g(x)) are located.

b) What is the horizontal asymptote as x approaches negative infinity of f(g(x))?

Jul 28, 2019
edited by Guest  Jul 28, 2019

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(1)    f(g (x) )  means :   put g  into f

Because the math is messy...I'm going to use WolframAlpha to simplify this  ...so f(g(x))  =

4(x - 5) (x - 2)            4 (x- 5)(x - 2)               4x^2 - 28x + 40

____________  =    ______________   =  ______________   (1)

3(x^2 - 6x - 7)           3 (x - 7) ( x + 1)            3x^2 - 18x - 21

The vertical asymptotes  are the x values that make the denominator  = 0  these are   x  = 7   and  x = -1

So....their  sum  =  7 + -1   =  6

(2) As x approaches pos/neg  infinity.....y approaches  the ratio of the coefficients on the x^2 terms in (1).....that is

y  =  4/3 ......and this is the horizontal asymptote  in both directions

See the graph here :  https://www.desmos.com/calculator/xg6a4ctnqz   Jul 28, 2019